pls help ive been stuck on this question for a while im not good with chemistry lol.

Answer:

Avogadro's number is 1 mol  = 6.02 * 10^23 elements

It means that 1 mol of atoms is 6.02 * 10^23 atoms

1 mol of atoms = 6.02 * 10^23 atoms

From there, if you divide both sides by 1 mol of atoms, you get

1 = 6.02 * 10^23 atoms / 1 mol of atoms.

That means, that to pass from a number of moles of atoms to number of atoms you have to multipby by the conversion factor

         6.02*10^23  atoms Al/ 1 mol Al

That is the second option of the list.

Explanation:

Answer:

The fluoride ion is capable of reacting, to a small extent, with water, accepting a proton. The fluoride ion is acting as a weak Brønsted-Lowry base. The hydroxide ion that is produced as a result of the above reaction makes the solution slightly basic.

So

we get a basic solution when you dissolve NaF in water.

Na is a alkaline earth metal

Metallic compound dissolved in water acts as base

Also there is another reason

Na F is ioniC

Fluorine is known as having highest electron affinity in world

It can accept a line pair of OH-

So it's Bronsted Lowry base .

Hence its basic

Answer:

H₂O

Explanation:

Two molecules of Hydrogen and one molecules of Oxygen, when mixed, create H₂O, or water. There is no scientific name for H₂O due to it's common name. It is just refereed to as "water" or H₂O.

Answer:

Part A

HCOOH(aq) + NaOH(aq) → HCOONa(aq) + H2O(l)

Part B

ClCH2CH2CO2H(aq) + NaOH(aq) ------> ClCH2CH2CO2Na(aq) + H2O(l)

Explanation:

The reaction between an alkanoic acid and a base is a neutralization reaction. The reaction occurs as follows;

RCOOH + NaOH ----> RCOONa + H2O

We have to note the fact that the net ionic reaction still remains;

H^+(aq) + OH^-(aq) ---> H2O(l)

In both cases, the reaction can occur and they actually do occur as written.

QUESTION:- Manganese-55 has _____neutrons.

OPTIONS :-

A. 55

B. 30

C. 25

ANSWER:- NUMBER OF NEUTRONS IS EQUAL TO THE DIFFERENCE BETWEEN THE MASS IF THE ATOM AND ATOMIC NUMBER

SO DIFFERENCE IS EQUAL TO :- 55-25 = 30 NEUTRONS.

SO THERE IS 30 NEUTRONS IN SINGLE ATOM OF THE MANGANESE-55 ATOM.

Answer:

the mass of an atom is the sum of proton and neutron which are both concentrated in nocleus of an atom. from the question the mass is given as 55 and the proton is 25.

Answer:

Element

Element : A pure substance composed of the same type of atom throughout. Compound : A substance made of two or more elements that are chemically combined in fixed amounts.

Explanation:

Answer:

0.0347 moles of hydronium ions

Explanation:

The equation of the neutralization reaction between hydroxide and hydronium ions is given below:

H₃O+ (aq) + OH- (aq) ----> 2 H₂O (l)

From the equation above, 1 mole of hydroxide ions will neutralize one mole hydronium ions.

The moles of hydroxide ions present in 1 teaspoon or 5 mL of antacid product is calculated as follows:

Number of moles = mass / molar mass

Molar mass of Magnesium hydroxide, Mg(OH)₂ = 58 g/mol

Molar mass of aluminium hydroxide, Al(OH)₃ = 78 g/mol

Mass of magnesium hydroxide = 450 g = 0.45 g

Mass of aluminium hydroxide = 500 mg = 0.5 g

Moles of magnesium hydroxide = (0.45/58) moles

Moles of aluminium hydroxide = (0.5/78) moles

Equation of the ionization of magnesium hydroxide and aluminium hydroxide is given below:

Mg(OH)₂ (aq) ----> Mg²+ (aq) + 2 OH- (aq)

Al(OH)₃ (aq) ---> Al³+ (aq) + 3 OH- (aq)

Number of moles of hydroxide ions present in (0.45/58) moles of magnesium hydroxide = 2 × (0.45/58) moles = 0.0155 moles

Number of moles of hydroxide ions present in (0.5/78) moles of aluminium hydroxide = 3 × (0.5/78) moles = 0.0192 moles

Total moles of hydroxide ions = 0.0155 + 0.0192 = 0.0347 moles hydroxide ions

Therefore, 0.0347 moles of hydroxide ions will neutralize 0.0347 moles of hydronium ions.

Answer:

17

Explanation:

Step 1: Calculate the needed concentrations

[A]i = 1.00 mol/5.00 L = 0.200 M

[B]i = 1.80 mol/5.00 L = 0.360 M

[B]e = 1.00 mol/5.00 L = 0.200 M

Step 2: Make an ICE chart

        A(aq) + 2 B(aq) ⇄ C(aq)

I       0.200    0.360        0

C        -x           -2x         +x

E     0.200-x  0.360-2x   x

Then,

[B]e = 0.360-2x = 0.200

x = 0.0800

The concentrations at equilibrium are:

[A]e = 0.200-0.0800 = 0.120 M

[B]e = 0.200 M

[C]e = 0.0800 M

Step 3: Calculate the concentration equilibrium constant (K)

K = [C] / [A] × [B]²

K = 0.0800 / 0.120 × 0.200² = 16.6 ≈ 17

Answer:

3.89 ×10^-5

Explanation:

Since they are gaseous reactants, we obtain the equilibrium constant from the given partial pressures;

p(H2O) = 0.0900 bar

p(H2) = 0.00100 bar

p(O2) = 0.00350 bar

The equation of the reaction is;2H2O(g)⇄2H2(g)+O2(g)

Kp= p(H2) . p(O2)/p(H2O)

Kp= 0.00100 × 0.00350/0.0900

Kp= 3.89 ×10^-5

Answer:

[tex]\boxed {\boxed {\sf 5.56 \times 10^{23} \ molecules \ CH_4}}[/tex]

Explanation:

We are asked to find how many molecules of methane are in 14.8 grams of the substance.

First, we convert grams to moles. We use the molar mass, or the mass of 1 mole of a substance. These values are equivalent to the atomic masses found on the Periodic Table, however the units are grams per mole instead of atomic mass units.

We are given the compound methane, or CH₄. Look up the molar mass of the individual elements (carbon and hydrogen).

Check the formula for subscripts. Hydrogen (H) has a subscript of 4, so there are 4 moles of hydrogen in 1 mole of methane. We must multiply hydrogen's molar mass by 4, then add carbon's molar mass.

Now we use dimensional analysis to convert. To do this, we set up a ratio using the molar mass.

[tex]\frac {16.043 \ g \ CH_4 }{ 1 \ mol \ CH_4}[/tex]

Since we are converting 14.8 grams of methane to moles, we multiply by this value.

[tex]14.8 \ g \ CH_4 *\frac {16.043 \ g \ CH_4 }{ 1 \ mol \ CH_4}[/tex]

Flip the ratio so the units of grams of methane cancel.

[tex]14.8 \ g \ CH_4 *\frac{ 1 \ mol \ CH_4} {16.043 \ g \ CH_4 }[/tex]

[tex]14.8 *\frac{ 1 \ mol \ CH_4} {16.043}[/tex]

[tex]\frac {14.8}{16.043} \ mol \ CH_4= 0.9225207256 \ mol \ CH_4[/tex]

Next, we convert moles to molecules. We use Avogadro's Number or 6.022  × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this problem, the particles are moles of methane. Set up another ratio using Avogadro's Number.

[tex]\frac { 6.022 \times \ 10^{23} \ molecules \ CH_4}{ 1 \ mol \ CH_4}[/tex]

Multiply by the number of moles we calculated.

[tex]0.9225207256\ mol \ CH_4 * \frac { 6.022 \times \ 10^{23} \ molecules \ CH_4}{ 1 \ mol \ CH_4}[/tex]

The units of moles of methane cancel.

[tex]0.9225207256* \frac { 6.022 \times \ 10^{23} \ molecules \ CH_4}{ 1 }[/tex]

[tex]5.55541981 \times 10^{23} \ molecules \ CH_4[/tex]

The original measurement of grams (14.8) has 3 significant figures, so our answer must have the same. For the number we calculated, that is the hundredth place. The 5 in the thousandth place tells us to round the 5 in the hundredth place up to a 6.

[tex]5.56 \times 10^{23} \ molecules \ CH_4[/tex]

14.8 grams of methane is equal to approximately 5.56 × 10²³ molecules of methane.

Answer:

0.0061650760770388 mole

Answer:

heat

Explanation:

because it's the cause of change

Answer:

heat

Explanation:

because it is a natural factor that causes the change in Earth's system

Answer:

Transition temperature is the temperature at which a substance changes from one state to another.

Allotropy is the existence of an element in many forms.

Answer:

D. An alkene ​

Explanation:

because Ethane is C2H4

Answer:

It's a alkANE. C.

Explanation:

The easiest way to memorize this is to look at the endings. Substances that end in -ANE are alkANEs. Substances that end in -ENE are alkENEs. Substances that end in -YNE are alkYNEs.

Answer:

A clinical thermometer is a thermometer used to measure human body temperature. Most made in the 20th century are mercury-in-glass thermometers. They are accurate and sensitive, having a narrow place where the mercury level rises very fast. A kink in the tube stops the mercury level from falling on its own.

Answer:

Explanation:

Start with a balanced equation.

2H2 + O2 → 2H2O

Calculate mole H2 using the formula: n = m/M, where:

n = mole

m = mass (g)

M = molar mass (g/mol)

Calculate molar mass of H2.

M H2 = 2 × 1.008 g/mol = 2.016 g/mol

Calculate moles H2.

n H2 = 4.16 g H2/2.016 g/mol = 2.063 mol H2

Calculate moles H2O by multiplying moles H2 by the mole ratio between H2O and H2 from the balanced equation, so that moles H2 cancel.

2.063 mol H2 × (2 mol H2O/2 mol H2) = 2.063 mol H2O

The mass of water will be calculated by rearranging the n = m/M formula to isolate m;

m = n × M

Calculate the molar mass H2O.

M H2O = (2 × 1.008 g/mol) + (1 × 15.999 g/mol) = 18.015 g/mol

Calculate the mass H2O.

m = n × M = 2.063 mol H2O × 18.015 g/mol = 37.2 g H2O

4.16 g H2 with excess O2 will produce 37.2 g H2O.

Answer:

As hydrogen is more electronegative than boron, in BH3 the oxidation number of hydrogen should be taken as -1. (E. N. of B = 2.0 & E. N. of H = 2.1)

Suppose, oxidation number of B is x.

So, we can write, x + 3*(-1) = 0

=> x = +3

Therefore, oxidation num

Explanation:

As hydrogen is more electronegative than boron, in BH3 the oxidation number of hydrogen should be taken as -1. (E. N. of B = 2.0 & E. N. of H = 2.1)

Suppose, oxidation number of B is x.

So, we can write, x + 3*(-1) = 0

=> x = +3

Therefore, oxidation num

2.8 g is the minimum mass of HCl in grams that would you need to dissolve a 2.2 g iron bar on a padlock.

When a solute is dissolved in a solvent, a solution is created. Dissolution is the process through which solutes, or dissolved parts, combine to form a solution inside a solvent. In this procedure, the gas, liquid, or solid dissolves inside the original solvent and forms a solution.

In some polymer applications, dissolution is also an issue since it results in swelling, a loss of strength and stiffness, and a change in volume. Whether a chemical process is man-made or natural, dissolution is crucial. Catalysts are tested using dissolution. 2.8 g is the minimum mass of HCl in grams that would you need to dissolve a 2.2 g iron bar on a padlock.

Therefore, 2.8 g is the minimum mass of HCl in grams that would you need to dissolve a 2.2 g iron bar on a padlock.

To know more about dissolution, here:

https://brainly.com/question/23851972

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Answer:

The correct answer is "654.54 nm".

Explanation:

According to the question,

⇒ [tex]\frac{1}{\lambda}= Rh(\frac{1}{n1^2} -\frac{1}{n2^2} )[/tex]

By substituting the values, we get

       [tex]=1.1\times 10^7(\frac{1}{4} -\frac{1}{9} )[/tex]

       [tex]=1.1\times 10^7(\frac{9-4}{36} )[/tex]

       [tex]=1.1\times 10^7(\frac{5}{36} )[/tex]

       [tex]=654.54\ nm[/tex]

Thus the above is the right solution.

Answer:

Answer is B) 20.0 moles

Explanation:

From the equation,

1 mole of Fe2O3 = 2 moles of Al

therefore 10.0 moles of Fe2O3 = 10×2

= 20.0 moles.

Answer:

C. yes, because it is universally applies to all objects

Answer:

However, there has to be 2 electrons on the first shell, and 8 on the others.

Explanation:

Hope this helps :)

Answer:

The lewis structure for NO₃⁻ is shown in the attachment below

For the central nitrogen atom:

The number of lone pairs = 0

The number of single bonds = 2  

The number of double bonds= 1

Explanation:

The lewis structure for NO₃⁻ is shown in the attachment below.

From the Lewis structure

For the central nitrogen atom:

The number of lone pairs = 0

The number of single bonds = 2  

The number of double bonds= 1

Answer:

C2H2 + Br2 → C2H2Br2

Explanation:

a) HNO₂ + C₂H₇NO → N₂ + C₂H₆O + H₂O

b) H₃O⁺ + F⁻ → HF + H₂O

[tex]\large\color{lime}\boxed{\colorbox{black}{Answer : - }}[/tex]

a) HNO₂ + C₂H₇NO → N₂ + C₂H₆O + H₂O

b) H₃O⁺ + F⁻ → HF + H₂O

Stoichiometry refers to the relationship between the moles of reactants and products.

This question must be solved using both stoichiometry and the gas laws

The reaction equation is;

2 Al(s) + 6 HCl(aq) --------> 2 AICI3, (aq) + 3 H2(g)

Number of moles of Al = 3g/27g/mol = 0.11 moles

According to the reaction equation;

2 moles of Al yields 3 moles of H2

0.11 moles of Al yields 0.11 * 3/2 = 0.165 moles

From the ideal gas equation;

PV=nRT

P = ?

n= 0.165 moles

V = 0.85 L

T = 297 K

R = 0.082 atmLK-1mol-1

P= nRT/V

P = 0.165 * 0.082 * 297/0.85

P= 4.73 atm

P1 = 4.73 atm

T1 = 297 K

V1 = 0.85 L

P2 = 1 atm

T2 =273 K

V2 =?

P1V1/T1 = P2V2/T2

P1V1T2 = P2V2T1

V2 = P1V1T2/P2T1

V2 =  4.73 * 0.85 * 273/1 * 297

V2 = 3.69 L

P1 = 760 mmHg

T1 = 273 K

V1 = 3.69

P2 = 438 mm Hg

T2 = 284.2 K

V2 =?

P1V1/T1 = P2V2/T2

P1V1T2 = P2V2T1

V2 = P1V1T2/P2T1

V2 = 760 * 3.69 * 284.2/438 *273

V2 = 797010.48/119574

V2= 6.67 L

https://brainly.com/question/1190311

Answer:

90.99 or 91.0

Explanation:

Using the balanced equation, you convert 38.5g of ethanol to moles of water. From there, you plug the values into the Ideal Gas Equation: PV=nRT.

Answer: The volume of oxygen gas is 91.4 L.

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)

Given mass of ethanol = 38.5 g

Molar mass of ethanol = 46 g/mol

Plugging values in equation 1:

[tex]\text{Moles of ethanol}=\frac{38.5g}{46g/mol}=0.840 mol[/tex]

The given chemical equation follows:

[tex]CH_3CH_2OH(g)+3O_2(g)\rightarrow 2CO_2(g)+3H_2O(g)[/tex]

By stoichiometry of the reaction:

If 1 mole of ethanol produces 3 moles of water

So, 0.840 moles of ethanol will produce = [tex]\frac{3}{1}\times 0.840=2.52mol[/tex] of water

The ideal gas equation is given as:

[tex]PV=nRT[/tex] .......(2)

where

P = pressure = 1.75 atm

V = volume of oxygen gas = ?

n = number of moles= 2.52 moles

R = Gas constant = 0.0821 L.atm/mol.K

T = temperature of the tank = [tex]500^oC=[500+273]K=773K[/tex]

Putting values in equation 2, we get:

[tex]1.75 atm\times V=2.52mol\times 0.0821L.atm/mol.K\times 773K\\\\V=\frac{2.52\times 0.0821\times 773}{1.75}=91.4L[/tex]

Hence, the volume of oxygen gas is 91.4 L.

Answer:

Explanation:

ok

Answer:

B

Explanation:

In order to create spontaneity, an endothermic process has to occur along with positive entropy and high temperature